# 009B Sample Final 1, Problem 4

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Compute the following integrals.

(a)  ${\displaystyle \int {\frac {t^{2}}{\sqrt {1-t^{6}}}}~dt}$

(b)  ${\displaystyle \int {\frac {2x^{2}+1}{2x^{2}+x}}~dx}$

(c)  ${\displaystyle \int \sin ^{3}x~dx}$

Foundations:
1. Through partial fraction decomposition, we can write the fraction
${\displaystyle {\frac {1}{(x+1)(x+2)}}={\frac {A}{x+1}}+{\frac {B}{x+2}}}$
for some constants ${\displaystyle A,B.}$
2. Recall the Pythagorean identity
${\displaystyle \sin ^{2}(x)+\cos ^{2}(x)=1.}$

Solution:

(a)

Step 1:
We first note that

${\displaystyle \int {\frac {t^{2}}{\sqrt {1-t^{6}}}}~dt=\int {\frac {t^{2}}{\sqrt {1-(t^{3})^{2}}}}~dt.}$

Now, we proceed by  ${\displaystyle u}$-substitution.
Let  ${\displaystyle u=t^{3}.}$
Then,   ${\displaystyle du=3t^{2}dt}$  and  ${\displaystyle {\frac {du}{3}}=t^{2}dt.}$
So, we have

${\displaystyle \int {\frac {t^{2}}{\sqrt {1-t^{6}}}}~dt=\int {\frac {1}{3{\sqrt {1-u^{2}}}}}~du.}$

Step 2:
Now, we need to use trig substitution.
Let  ${\displaystyle u=\sin \theta .}$  Then,  ${\displaystyle du=\cos \theta d\theta .}$
So, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {t^{2}}{\sqrt {1-t^{6}}}}~dt}&=&\displaystyle {\int {\frac {\cos \theta }{3{\sqrt {1-\sin ^{2}\theta }}}}~d\theta }\\&&\\&=&\displaystyle {\int {\frac {\cos \theta }{3{\sqrt {\cos ^{2}\theta }}}}~d\theta }\\&&\\&=&\displaystyle {\int {\frac {\cos \theta }{3\cos \theta }}d\theta }\\&&\\&=&\displaystyle {\int {\frac {1}{3}}~d\theta }\\&&\\&=&\displaystyle {{\frac {1}{3}}\theta +C}\\&&\\&=&\displaystyle {{\frac {1}{3}}\arcsin(u)+C}\\&&\\&=&\displaystyle {{\frac {1}{3}}\arcsin(t^{3})+C.}\end{array}}}$

(b)

Step 1:
First, we add and subtract  ${\displaystyle x}$  from the numerator.
So, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx}&=&\displaystyle {\int {\frac {2x^{2}+x-x+1}{2x^{2}+x}}~dx}\\&&\\&=&\displaystyle {\int {\frac {2x^{2}+x}{2x^{2}+x}}~dx+\int {\frac {1-x}{2x^{2}+x}}~dx}\\&&\\&=&\displaystyle {\int 1~dx+\int {\frac {1-x}{2x^{2}+x}}~dx}.\\\end{array}}}$
Step 2:
Now, we need to use partial fraction decomposition for the second integral.
Since  ${\displaystyle 2x^{2}+x=x(2x+1),}$  we let
${\displaystyle {\frac {1-x}{2x^{2}+x}}={\frac {A}{x}}+{\frac {B}{2x+1}}.}$
Multiplying both sides of the last equation by  ${\displaystyle x(2x+1),}$
we get
${\displaystyle 1-x=A(2x+1)+Bx.}$
If we let  ${\displaystyle x=0,}$ the last equation becomes  ${\displaystyle 1=A.}$
If we let  ${\displaystyle x=-{\frac {1}{2}},}$  then we get  ${\displaystyle {\frac {3}{2}}=-{\frac {1}{2}}\,B.}$  Thus,  ${\displaystyle B=-3.}$
So, in summation, we have
${\displaystyle {\frac {1-x}{2x^{2}+x}}={\frac {1}{x}}+{\frac {-3}{2x+1}}.}$
Step 3:
If we plug in the last equation from Step 2 into our final integral in Step 1, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx}&=&\displaystyle {\int 1~dx+\int {\frac {1}{x}}~dx+\int {\frac {-3}{2x+1}}~dx}\\&&\\&=&\displaystyle {x+\ln x+\int {\frac {-3}{2x+1}}~dx.}\\\end{array}}}$

Step 4:
For the final remaining integral, we use  ${\displaystyle u}$-substitution.
Let  ${\displaystyle u=2x+1.}$
Then,  ${\displaystyle du=2\,dx}$  and  ${\displaystyle {\frac {du}{2}}=dx.}$
Thus, our integral becomes

${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx}&=&\displaystyle {x+\ln x+\int {\frac {-3}{2x+1}}~dx}\\&&\\&=&\displaystyle {x+\ln x+\int {\frac {-3}{2u}}~du}\\&&\\&=&\displaystyle {x+\ln x-{\frac {3}{2}}\ln u+C.}\\\end{array}}}$

${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx}&=&\displaystyle {x+\ln x-{\frac {3}{2}}\ln(2x+1)+C.}\end{array}}}$

(c)

Step 1:
First, we write
${\displaystyle \int \sin ^{3}x~dx=\int \sin ^{2}x\sin x~dx.}$
Using the identity  ${\displaystyle \sin ^{2}x+\cos ^{2}x=1,}$  we get
${\displaystyle \sin ^{2}x=1-\cos ^{2}x.}$
If we use this identity, we have
${\displaystyle \int \sin ^{3}x~dx=\int (1-\cos ^{2}x)\sin x~dx.}$
Step 2:
Now, we proceed by  ${\displaystyle u}$-substitution.
Let  ${\displaystyle u=\cos x.}$  Then,  ${\displaystyle du=-\sin xdx.}$
So we have

${\displaystyle {\begin{array}{rcl}\displaystyle {\int \sin ^{3}x~dx}&=&\displaystyle {\int -(1-u^{2})~du}\\&&\\&=&\displaystyle {-u+{\frac {u^{3}}{3}}+C}\\&&\\&=&\displaystyle {-\cos x+{\frac {\cos ^{3}x}{3}}+C}.\\\end{array}}}$

(a)    ${\displaystyle {\frac {1}{3}}\arcsin(t^{3})+C}$
(b)    ${\displaystyle x+\ln x-{\frac {3}{2}}\ln(2x+1)+C}$
(c)    ${\displaystyle -\cos x+{\frac {\cos ^{3}x}{3}}+C}$